Fitting Exponential Models to Data

In previous sections of this chapter, we were either given a function explicitly to graph or evaluate, or we were given a set of points that were guaranteed to lie on the curve. Then we used algebra to find the equation that fit the points exactly. In this section, we use a modeling technique called regression analysis to find a curve that models data collected from real-world observations. With regression analysis, we don’t expect all the points to lie perfectly on the curve. The idea is to find a model that best fits the data. Then we use the model to make predictions about future events.

Do not be confused by the word model. In mathematics, we often use the terms function, equation, and model interchangeably, even though they each have their own formal definition. The term model is typically used to indicate that the equation or function approximates a real-world situation.

We will concentrate on three types of regression models in this section: exponential, logarithmic, and logistic. Having already worked with each of these functions gives us an advantage. Knowing their formal definitions, the behavior of their graphs, and some of their real-world applications gives us the opportunity to deepen our understanding. As each regression model is presented, key features and definitions of its associated function are included for review. Take a moment to rethink each of these functions, reflect on the work we’ve done so far, and then explore the ways regression is used to model real-world phenomena.

Building an Exponential Model from Data

As we’ve learned, there are a multitude of situations that can be modeled by exponential functions, such as investment growth, radioactive decay, atmospheric pressure changes, and temperatures of a cooling object. What do these phenomena have in common? For one thing, all the models either increase or decrease as time moves forward. But that’s not the whole story. It’s the way data increase or decrease that helps us determine whether it is best modeled by an exponential equation. Knowing the behavior of exponential functions in general allows us to recognize when to use exponential regression, so let’s review exponential growth and decay.

Recall that exponential functions have the form $y=ab^x$ or $y=A_0e^{kx}$. When performing regression analysis, we use the form most commonly used on graphing utilities, $y=ab^x$. Take a moment to reflect on the characteristics we’ve already learned about the exponential function $y=ab^x$ (assume $a>0$):

• $b$ must be greater than zero and not equal to one.
• The initial value of the model is $y=a$.
  • If $b>1$, the function models exponential growth. As $x$ increases, the outputs of the model increase slowly at first, but then increase more and more rapidly, without bound.
  • If $0<b<1$, the function models exponential decay. As $x$ increases, the outputs for the model decrease rapidly at first and then level off to become asymptotic to the $x$-axis. In other words, the outputs never become equal to or less than zero.

As part of the results, your calculator will display a number known as the correlation coefficient, labeled by the variable $r$, or $r^2$. (You may have to change the calculator’s settings for these to be shown.) The values are an indication of the “goodness of fit” of the regression equation to the data. We more commonly use the value of $r^2$ instead of $r$, but the closer either value is to 1, the better the regression equation approximates the data.

Example. In 2007, a university study was published investigating the crash risk of alcohol impaired driving. Data from 2,871 crashes were used to measure the association of a person’s blood alcohol level (BAC) with the risk of being in an accident. The table below shows results from the study. The relative risk is a measure of how many times more likely a person is to crash. So, for example, a person with a BAC of $0.09$ is $3.54$ times as likely to crash as a person who has not been drinking alcohol.

1. Let $x$ represent the BAC level, and let $y$ represent the corresponding relative risk. Use exponential regression to fit a model to these data.
2. After 6 drinks, a person weighing 160 pounds will have a BAC of about $0.16$. How many times more likely is a person with this weight to crash if they drive after having a 6-pack of beer? Round to the nearest hundredth.

Solution.

1. Using the STAT then EDIT menu on a graphing utility, list the BAC values in L1 and the relative risk values in L2. Then use the STATPLOT feature to verify that the scatterplot follows the exponential pattern shown below:

   Use the “ExpReg” command from the STAT then CALC menu to obtain the exponential model,

   $$y=0.58304829(2.20720213\text{E}10)^x$$

   Converting from scientific notation, we have:

   $$y=0.58304829(22{,}072{,}021{,}300)^x$$

   Notice that $r^2\approx0.97$ which indicates the model is a good fit to the data. To see this, graph the model in the same window as the scatterplot to verify it is a good fit, as shown below:

2. Use the model to estimate the risk associated with a BAC of $0.16$. Substitute $0.16$ for $x$ in the model and solve for $y$.

   $$\begin{array}{lrcl} & y &=& 0.58304829(22{,}072{,}021{,}300)^x \\[4pt] \text{Substitute }0.16\text{ for }x. & &=& 0.58304829(22{,}072{,}021{,}300)^{0.16} \\[4pt] \text{Round to the nearest hundredth.} & &\approx& 26.35 \end{array}$$

   If a 160-pound person drives after having 6 drinks, they are about $26.35$ times more likely to crash than if driving while sober.

The table below shows a recent graduate’s credit card balance each month after graduation.

Building a Logarithmic Model from Data

Just as with exponential functions, there are many real-world applications for logarithmic functions: intensity of sound, pH levels of solutions, yields of chemical reactions, production of goods, and growth of infants. As with exponential models, data modeled by logarithmic functions are either always increasing or always decreasing as time moves forward. Again, it is the way they increase or decrease that helps us determine whether a logarithmic model is best.

Recall that logarithmic functions increase or decrease rapidly at first, but then steadily slow as time moves on. By reflecting on the characteristics we’ve already learned about this function, we can better analyze real world situations that reflect this type of growth or decay. When performing logarithmic regression analysis, we use the form of the logarithmic function most commonly used on graphing utilities, $y=a+b\ln(x)$. For this function

• All input values, $x$, must be greater than zero.
• The point $(1,a)$ is on the graph of the model.
• If $b>0$, the model is increasing. Growth increases rapidly at first and then steadily slows over time.
• If $b<0$, the model is decreasing. Decay occurs rapidly at first and then steadily slows over time.

Example. Due to advances in medicine and higher standards of living, life expectancy has been increasing in most developed countries since the beginning of the 20th century.

The table below shows the average life expectancies, in years, of Americans from 1900–2010.

1. Let $x$ represent time in decades starting with $x=1$ for the year 1900, $x=2$ for the year 1910, and so on. Let $y$ represent the corresponding life expectancy. Use logarithmic regression to fit a model to these data.
2. Use the model to predict the average American life expectancy for the year 2030.

Solution.

1. Using the STAT then EDIT menu on a graphing utility, list the years using values 1–12 in L1 and the corresponding life expectancy in L2. Then use the STATPLOT feature to verify that the scatterplot follows a logarithmic pattern, as shown below:

   Use the “LnReg” command from the STAT then CALC menu to obtain the logarithmic model,

   $$y=42.52722583+13.85752327\ln(x)$$

   Next, graph the model in the same window as the scatterplot to verify it is a good fit, as shown below:

2. To predict the life expectancy of an American in the year 2030, substitute $x=14$ for the decade in the model and solve for $y$:

   $$\begin{array}{lrcl} & y &=& 42.52722583+13.85752327\ln(x) \\[4pt] \text{Substitute }14\text{ for }x. & &=& 42.52722583+13.85752327\ln(14) \\[4pt] \text{Round to the nearest tenth.} & &\approx& 79.1 \end{array}$$

   If life expectancy continues to increase at this pace, the average life expectancy of an American will be $79.1$ by the year 2030.

Sales of a video game released in the year 2000 took off at first, but then steadily slowed as time moved on. The table below shows the number of games sold, in thousands, from the years 2000–2010.

Building a Logistic Model from Data

Like exponential and logarithmic growth, logistic growth increases over time. One of the most notable differences with logistic growth models is that, at a certain point, growth steadily slows and the function approaches an upper bound, or limiting value. Because of this, logistic regression is best for modeling phenomena where there are limits in expansion, such as availability of living space or nutrients.

It is worth pointing out that logistic functions actually model resource-limited exponential growth. There are many examples of this type of growth in real-world situations, including population growth and spread of disease, rumors, and even stains in fabric. When performing logistic regression analysis, we use the form most commonly used on graphing utilities:

Recall that:

• $\tfrac{c}{1+a}$ is the initial value of the model.
• when $b>0$, the model increases rapidly at first until it reaches its point of maximum growth rate, $\left(\tfrac{\ln(a)}{b},\tfrac{c}{2}\right)$. At that point, growth steadily slows and the function becomes asymptotic to the upper bound $y=c$.
• $c$ is the limiting value, sometimes called the carrying capacity, of the model.

Example. Mobile telephone service has increased rapidly in America since the mid 1990s. Today, almost all residents have cellular service. The table below shows the percentage of Americans with cellular service between the years 1995 and 2012.

1. Let $x$ represent time in years starting with $x=0$ for the year 1995. Let $y$ represent the corresponding percentage of residents with cellular service. Use logistic regression to fit a model to these data.
2. Use the model to calculate the percentage of Americans with cell service in the year 2013. Round to the nearest tenth of a percent.
3. Discuss the value returned for the upper limit, $c$. What does this tell you about the model? What would the limiting value be if the model were exact?

Solution.

1. Using the STAT then EDIT menu on a graphing utility, list the years using values 0–17 in L1 and the corresponding percentage in L2. Then use the STATPLOT feature to verify that the scatterplot follows a logistic pattern, as shown below:

   Use the “Logistic” command from the STAT then CALC menu to obtain the logistic model,

   $$y=\frac{105.7379526}{1+6.88328979e^{-0.2595440013x}}$$

   Next, graph the model in the same window as the scatterplot to verify it is a good fit, as shown below:

2. To approximate the percentage of Americans with cellular service in the year 2013, substitute $x=18$ for the year in the model and solve for $y$:

   $$\begin{array}{lrcl} & y &=& \tfrac{105.7379526}{1+6.88328979e^{-0.2595440013x}} \\[8pt] \text{Substitute }18\text{ for }x. & &=& \tfrac{105.7379526}{1+6.88328979e^{-0.2595440013(18)}} \\[8pt] \text{Round to the nearest tenth.} & &\approx& 99.3 \end{array}$$

   According to the model, about $99.3\%$ of Americans had cellular service in 2013.

3. The model gives a limiting value of about $105$. This means that the maximum possible percentage of Americans with cellular service would be $105\%$, which is impossible. (How could over $100\%$ of a population have cellular service?) If the model were exact, the limiting value would be $c=100$ and the model’s outputs would get very close to, but never actually reach $100\%$. After all, there will always be someone out there without cellular service!

The table below shows the population, in thousands, of harbor seals in the Wadden Sea over the years 1997 to 2012.

Key concepts

• Exponential regression is used to model situations where growth begins slowly and then accelerates rapidly without bound, or where decay begins rapidly and then slows down to get closer and closer to zero.
• We use the command “ExpReg” on a graphing utility to fit function of the form $y=ab^x$ to a set of data points.
• Logarithmic regression is used to model situations where growth or decay accelerates rapidly at first and then slows over time.
• We use the command “LnReg” on a graphing utility to fit a function of the form $y=a+b\ln(x)$ to a set of data points.
• Logistic regression is used to model situations where growth accelerates rapidly at first and then steadily slows as the function approaches an upper limit.
• We use the command “Logistic” on a graphing utility to fit a function of the form $y=\tfrac{c}{1+ae^{-bx}}$ to a set of data points.

Practice

Build an exponential model from data

The table below shows a set of data.

The table below shows a different set of data.

Build a logarithmic model from data

The table below shows a set of data.

Build a logistic model from data

The population $P$ of a koi pond over $x$ months is modeled by the function $P(x)=\tfrac{68}{1+16e^{-0.28x}}$.

The population $P$ of an endangered species habitat for wolves is modeled by the function $P(x)=\tfrac{558}{1+54.8e^{-0.462x}}$, where $x$ is given in years.

This section is adapted from Precalculus 2e, Section 4.8: Fitting Exponential Models to Data by Jay Abramson and OpenStax, © OpenStax, licensed under CC BY-NC-SA 4.0. Access the original for free at openstax.org. Changes: recreated every scatter plot and fitted-curve graph as an accessible inline SVG built from its exact data and equation — the BAC/relative-risk scatter and its exponential fit $y=0.58304829(22{,}072{,}021{,}300)^x$; the life-expectancy scatter and its logarithmic fit $y=42.52722583+13.85752327\ln(x)$; and the cellular-service scatter and its logistic fit $y=\tfrac{105.7379526}{1+6.88328979e^{-0.2595440013x}}$, whose curve has no closed-form primitive in this site’s figure engine and is therefore drawn as a dense polyline sampled directly from that exact equation rather than any spline approximation. Every regression coefficient printed on this page was independently re-derived — ordinary least squares of $\ln y$ against $x$ for each exponential model, of $y$ against $\ln x$ for each logarithmic model, and Nelder–Mead nonlinear least squares for each logistic model — and matched the source’s published values to the precision shown. Omitted the section’s one Media link to an external graphing-calculator resource, which carries no transcribable mathematics. Converted the three in-page Try Its (credit-card debt, video-game sales, and harbor-seal population) into interactive fill-in components, each part of a multipart Try It becoming its own component; converted eleven selected end-of-section exercises — two exponential-model builds from data tables, one logarithmic-model build plus two logarithmic evaluations, and six logistic-model evaluations across three scenarios (a generic carrying- capacity model, a koi pond, and a wolf habitat) — into interactive fill-ins in a closing Practice block, one group per objective. Two exercises whose end-of-section items depend on a graphing-calculator “intersect” reading were given the underlying model directly in the question text so each is answerable from the page alone. One upstream defect is corrected here: the pinned CNXML’s solution to the Table 9 logarithmic-regression exercise (module m49368, exercise id fs-id1650282) prints its fitted model as $y=5.063+1.934\log(x)$, but the section’s own “LnReg” definition returns $y=a+b\ln(x)$, and reproducing the regression independently on the printed data confirms the natural logarithm, not the common logarithm — this page’s Practice block states the model with $\ln(x)$.